Pai Gow Poker Computations

Brian Alspach

31 July 2003

Abstract:

This file provides the details for the numbers and probabilities given in the article ``Pai Gow Poker'' that appeared in Poker Digest, Vol. 5, No. 7, March 22 - April 4, 2002.

As a first step, let us list the hands we wish to count. In descending order of likelihood of occurrence they are:

The dollar figure shown in parentheses after each hand is the amount you are paid for a $1 bonus bet if you are dealt the particular hand.

The game is dealt from a standard 52-card deck together with a single joker. Thus, the total number of hands is . The basic approach we take is to partition the hands into those containing a joker and those not containing a joker. For convenience we now give tables of 7-card hands and 6-card hands for a standard 52-card deck. The derivation of these tables may be found elsewhere in the directory containing this file.

Table 3 below gives the number of various hands in which we are interested. We now begin the development of filling in that table.

The hands with no joker correspond to 7-card hands, so we first look at the 7-card hands to see where they contribute to the hands in Table 3. The important feature of 7-card hands is that, other than straight flushes, they contribute only to the categories to which they already belong. Therefore, we need consider only the 41,584 straight flushes. We must partition them into 7-card straight flushes without joker, royal flushes with royal match, royal flushes, and straight flushes.

It is easy to see there are 32 cards with which a 7-card straight flush may begin. This is the only way we can get a 7-card straight flush without a joker giving us the first entry of Table 3.

There are 4 choices of suit for a royal flush and choices for the other 2 cards. Of the latter choices, exactly 1 gives a 7-card straight flush without joker and they have been counted. Since there are 3 suited K,Q combinations from other suits, exactly 3 choices give a royal flush with royal match making a contribution of 12 towards that category. The remaining 1,077 choices produce 4,308 royal flushes not using a joker. This leaves 37,232 contributions to straight flushes.

In summary, the 41,584 7-card hands with a straight flush contribute

to Table 3.

This brings us to an examination of 6-card hands to determine the effect of adding a joker. We must be somewhat careful here as it is easy to overlook hands, and the addition of a joker can make a big difference. We also will make use of the file entitled Rank Sets and Straights found in the same directory containing this file. It has some useful information regarding sets of ranks that allow straights upon adding a joker. This is the area most liable to produce errors in computations about what happens when a joker is added to 6-card hands.

First we consider the 1,844 straight flushes. Of these hands, 36 consist of 6 suited cards of 6 consecutive ranks. Adding a joker to them makes a 7-card straight flush with joker putting 36 in that category. There are 1,808 with 5 consecutive ranks. The 4 suited rank sets {A,2,3,4,5}, {2,3,4,5,6}, {9,10,J,Q,K}, and {10,J,Q,K,A} have exactly 1 additional card such that adding a joker gives a 7-card straight flush with joker. The other sets of 5 consecutive suited cards allow 2 additional cards contributing to this category. That gives us 64 more hands producing a 7-card straight flush with joker. Altogether, we have picked up 100 in the latter category.

The suited rank set {10,J,Q,K,A} has 45 choices not yet considered. Adding the joker changes nothing and we have 180 hands which are royal flushes. The suited rank set {9,10,J,Q,K} has 44 choices not yet considered. The addition of a joker moves these hands to royal flushes giving us 176 royal flushes. Thus, we have 356 contributions to royal flushes.

All of the remaining 1,388 straight flushes remain just straight flushes upon adding the joker. In summary, the 1,844 6-card straight flush hands contribute

to Table 3.

There are 14,664 4-of-a-kind 6-card hands. Of these, 1,128 have 4 aces and upon adding a joker, they become 5 aces. This is the only way to get the hand 5 aces thereby verifying the 4th entry of Table 3. The other 13,536 4-of-a-kind hands remain as 4-of-a-kind upon adding a joker. Summarizing, the 14,664 4-of-a-kind 6-card hands contribute

to Table 3.

We now move to the 165,984 6-card hands with full houses. Of these, 12,864 contain 3 aces. They become 4 aces upon adding a joker and contribute to the 4-of-a-kind total. All of the remaining 153,120 hands, including those with 2 aces, remain full houses upon adding a joker. In summary, 165,984 full house 6-card hands contribute

to Table 3.

There are 205,792 6-card hands that are flushes. Note that such a hand has at least 5 distinct ranks so that the addition of a joker may not improve the hand to a full house or quads; that is, if a hand improves, it must become some kind of straight flush.

First consider the 6,580 flushes with 6 suited cards. This means there are 6 ranks present and there are sets of 6 ranks. There are 9 sets of 6 consecutive ranks and 62 sets with 5 consecutive ranks, all of which have been considered earlier. This leaves 1,645 rank sets without 5 or 6 consecutive ranks. We must determine the rank sets that produce straight flushes upon adding a joker.

First, there are 8 rank sets of the form

It is possible to obtain some of the other straight flushes, but impossible to get a royal flush with a royal match when the 6-card hand has 6 suited cards. If the ranks 10,J,Q,K are present, then we are choosing the other 2 ranks from {2,3,...,8}, but not allowed to choose 7,8. This yields 20 choices. If we have any of {10,J,Q,A}, {10,J,K,A}, {10,Q,K,A} or {J,Q,K,A}, then we are choosing 2 ranks from {2,3,...,9} except for 8,9. This gives us 27 choices. Altogether, we have 128 choices and multiplying by 4 yields 512 6-card hands with 6 suited cards that produce royal flushes upon adding a joker.

This leaves us the determination of the other hands with 6 suited cards that produce a straight flush upon adding a joker. There are 9 choices of 4 consecutive ranks of the form x,x+1,x+2,x+3 since x can take on any of the values A through 9, whereas, x = 10 and x = J already have been considered. They do not behave uniformly so let us describe the different behaviors.

A,2,3,4 allows us to choose any 2 from ranks 6 through K except 6,7. This yields 27 choices. The set {2,3,4,5} allows us to choose any 2 from ranks 7 through K except 7,8. This yields 20 choices. The ranks 3,4,5,6 and 9,10,J,Q behave similarly, except that we cannot choose either K or A for the latter set, giving us 20 choices for 3,4,5,6 and 14 choices for 9,10,J,Q. The other 5 sets allow only 19 choices because there are 2 excluded subsets of cardinality 2. Altogether, we have 176 choices which is multiplied by 4 to give us 704 straight flushes from 6 suited cards having 4 consecutive ranks.

Now consider sets of 6 ranks not having 4 consecutive ranks. We are going to count these according to the largest straight flush that may occur. If 4 is removed from the rank set {A,2,3,4,5}, then the other 2 ranks are chosen from 7 through K (including a 6 gives a bigger straight flush upon adding a joker) giving 21 sets. If 3 is removed, then the other 2 ranks are chosen from 7 through K except for the choice of 7,8. This produces another 20 sets. If 2 is removed, then we are choosing any 2 elements from 8 through K. This gives us another 15 sets or 56 sets in total. The rest of the sets behave similarly.

If 5 is removed from the rank set {2,3,4,5,6}, then the other 2 ranks are chosen from 8 through K giving 15 choices. When 4 is removed, then we are choosing 2 ranks from 8 through A, because including A does not produce 4 consecutive ranks, but we cannot choose both 8,9. Hence, there are 20 choices. Finally, when 3 is removed, we may not use 8 and are choosing 2 ranks from 9 through A giving 15 choices for a total of 50 choices altogether.

For the rank set {3,4,5,6,7}, when 6 is removed, we choose 2 ranks from 9 through A giving 15 choices. When 5 is removed, we choose 2 from 2,9 through A, except A,2 and 9,10. This gives 19 choices. Finally, when 4 is removed, we are choosing 2 from 2,10 through A, except A,2 itself, leading to another 14 choices. This gives 48 altogether.

The 4 sets of this type with smallest element 4, 5, 6 or 7 behave similarly with 48 sets. The set with smallest element 8 has 30 sets because {8,9,J,Q} produces 20 sets with an excluded pair no longer existing. This gives us 49 sets here. Finally, the set with smallest element 9 has 55 sets as can be checked easily.

Adding all of these gives 450 sets of 6 ranks without 4 consecutive ranks that produce 5 consecutive ranks upon adding a joker. This is multiplied by 4 to give 1,800 straight flushes.

This leaves 3,468 flushes with 6 suited cards that do not change upon adding a joker. In summary, the 6,580 6-card flushes contribute

We now move to the 199,212 6-card hands that are flushes and have 5 suited cards. Since there are only 5 suited cards in all of these hands, we never obtain any of the straight flushes of length 7 upon adding a joker. Thus, we are interested only in the 5 ranks used to form the flush. The 6th card may be any of the 39 cards not from the flush suit.

For those that become straight flushes with a joker, we classify them according to the rank of the straight flush. We already have examined 6-card hands with a straight flush so that we are concerned with the 1,277 rank sets without 5 consecutive ranks.

The only way we obtain a royal flush is from one of the subsets of the rank set {10,J,Q,K,A}. There are 7 choices of the remaining rank for the set {10,J,Q,K} and 39 choices for the other card giving us royal flushes in this case. For each of the 4 subsets obtained by deleting one of 10, J, Q or K, the 5th rank can be any of 2 through 9. In these cases, we then have royal flushes giving us 6,084 royal flushes arising from 6-card hands with 5 suited cards.

For an (x+4)-high straight flush, our set of ranks must contain 4 ranks from the set {x,x+1,x+2,x+3,x+4}, where x runs through A,2,...,9. The subset {x+1,x+2,x+3,x+4} never works because this leads to an (x+5)-high straight flush. If we delete the rank x+4, then we may not use x+5 or x-1 since the former allows a bigger straight flush and the latter introduces 5 consecutive ranks. As long as x is not A, this gives us (x+4)-high straight flushes. When x = A, we obtain 5-high straight flushes. We obtain 8,580 straight flushes from these sets.

For {x,x+1,x+2,x+3,x+4}, if we delete either x+2 or x+3, there are 7 choices for the rank of the 5th card of the suit. This yields straight flushes in this case. Running over the 9 values of x, we obtain another 19,656 straight flushes.

When we remove x+1, we cannot use rank x+5 or x+6. The rank x+6 does not make sense when x is 9. So we obtain 8,580 straight flushes in this subcase.

Summing gives 36,816 straight flushes that are not royal flushes. Thus the contributions to Table 3 are

We now arrive at straights. A 6-card hand with a straight has either 5 or 6 distinct ranks in the hand. Thus, the only improvement possible upon adding a joker is for a hand to become some kind of flush. This will take place when there are precisely 4 suited cards among the 6 cards in the hand. We break the consideration of these hands into 2 subcases depending on the number of distinct ranks in the hand. We consider hands with 6 distinct ranks first.

For a given rank set with 6 distinct ranks, there are possible hands. Exactly 4 of them consist of 6 suited cards. There are 72 of them consisting of 5 suited cards. The number of hands with exactly 4 suited cards is given by

There are 9 rank sets with 6 consecutive ranks, and 62 sets of 6 distinct ranks 5 of which are consecutive. Consider the given rank set {x,x+1,...,x+5}. Clearly, a straight flush arises unless both x and x+5 are in the flush suit. Of the 15 ways of choosing which 4 ranks will be suited, 6 of them have both x and x+5 in the suit. Thus, the other 9 choices give rise to a straight flush producing 324 straight flushes and 216 flushes for each of the 9 rank suits. In the special case that the rank set is {9,10,J,Q,K,A}, 5 of the choices for the 4 ranks to be suited lead to a royal flush. We get 180 royal flushes. Similarly, the rank set {8,9,10,J,Q,K} has 1 choice of 4 suited cards leading to a royal flush so that 36 royal flushes arise for this set. The summary for the 9 rank sets with 6 consecutive ranks is as follows: 216 royal flushes, 2,700 straight flushes, 1,944 flushes and 31,320 straights upon adding a joker to the 6-card hands.

Now consider the 62 sets with 6 distinct ranks, where 5 of them are consecutive. As above, each such set produces 540 hands with exactly 4 suited cards. There are several patterns of straight flush behavior depending on the distribution of ranks. Suppose the ranks have the form x,x+1,x+2,x+3,x+4,y, where y has a gap of at least 2 from either x or x+4. Then there are 5 choices of 4 ranks leading to a straight flush when a joker is added, and 10 choices of 4 ranks leading to a flush. Thus, of the 540 hands, 180 produce straight flushes and 360 produce flushes.

If y has a gap of only 1, then there are 6 choices of ranks leading to a straight flush and only 9 leading to a flush. This kind of rank set then produces 216 straight flushes and 324 flushes.

The special case of {10,J,Q,K,A} leads to 180 royal flushes among its straight flushes, and the special case of {9,10,J,Q,K} has 36 royal flushes from its collection of straight flushes. The rank set {8,9,10,J,Q,A} also contributes 36 royal flushes and 180 straight flushes.

Once we have worked out the number of rank sets with the various distribution patterns, we can determine the number of flushes. The rank set {A,2,3,4,5} has 1 value of y with a gap of 1 and 6 values of y with a gap of 2 or more. The rank set {2,3,4,5,6} has 1 value of y with a gap of 1 and 5 values of y with a gap of 2 or more. The 6 rank sets with smallest rank from 3 through 8 have 2 values of y with a gap of 1 and 4 values of y with a gap of at least 2. The rank set {9,10,J,Q,K} has 1 value of y with a gap of 1 and 5 values of y with a gap of at least 2. Finally, the rank set {10,J,Q,K,A} has 1 value of y with a gap of 1 and 6 values of y with a gap of at least 2. Altogether we have 16 of the rank sets with a gap of 1, and 46 rank sets with a gap of at least 2. However, of the 16 with a gap of 1, 1 of them involves {9,10,J,Q,K}, 1 of them involves {9,10,J,Q,A}, and 1 of them involves {10,J,Q,K,A}. Of the 46 with gap at least 2, 5 involve {9,10,J,Q,K} and 6 involve {10,J,Q,K,A}.

We now perform the arithmetic for the preceding rank sets. The rank sets {8,9,10,J,Q,A} and {7,9,10,J,Q,K} each contribute 36 royal flushes and 180 straight flushes. The rank set {8,10,J,Q,K,A} produces 180 royal flushes, 36 straight flushes, and 324 flushes. The 5 rank sets for {9,10,J,Q,K} with gap at least 2 produce 180 royal flushes, 720 straight flushes, and 1,800 flushes. The 6 rank sets for {10,J,Q,K,A} with gap at least 2 produce 1,080 royal flushes and 2,160 flushes. The remaining 13 rank sets with gap 1 produce 2,808 straight flushes and 4,212 flushes. The other 35 rank sets with gap at least 2 produce 6,300 straight flushes and 12,600 flushes. The grand totals then are 1,512 royal flushes, 10,224 straight flushes, 21,744 flushes, and 215,760 straights coming from the 62 rank sets of 6 distinct ranks with 5 consecutive ranks.

The last category of straights to consider are those that have 5 distinct ranks. The only option is for the 5 ranks to be consecutive and there are 10 such rank sets. There are 5 choices for which of the ranks are suited and 4 choices for the suit. If the pair occurs in the non-suited rank, then there are 3 choices for the pair. If the pair occurs in one of the suited ranks, there are 4 choices for the rank, 3 choices for the card, and 3 choices for the non-suited card in the 5 consecutive ranks. This gives us 39 ways to complete the 4 suited cards to include a pair, thereby, giving us 780 hands with 4 suited cards for the given rank set. For every one of these hands, the addition of a joker produces a straight flush. Therefore, we have 7,800 straight flushes arising. However, for the rank set {10,J,Q,K,A}, all 780 straight flushes are, in fact, royal flushes. In addition, some of these royal flushes involve a royal match, that is, a suited K,Q in a suit other than the suit of the royal flush. When we have 10,J,Q,A suited and a K,Q from another suit, the joker then provides us with a royal flush with a royal match. There are 12 ways for this to happen. We also get 12 of them when the 10,J,K,A are suited.

In addition, for the rank set {9,10,J,Q,K}, 156 of the straight flushes are royal flushes. Thus, we end up with 24 royal flushes with a royal match, 912 royal flushes, 6,864 straight flushes and 68,400 straights after adding a joker.

Summing all of the outcomes for 6-card straights, the contributions to Table 3 are

We now consider the 732,160 3-of-a-kind 6-card hands (affectionately known as trips). Since 4 ranks are present, the addition of a joker can change such a hand in many ways. There are 56,320 of these hands with the form A,A,A,x,y,z. The addition of a joker produces at worst 4 aces. We need to separate those that become some kind of straight flush. So if x,y,z is chosen from {2,3,4,5}, some of the hands will become a straight flush. There are 4 choices for the 3 ranks, 4 choices for the common suit of the 3 ranks, and 3 choices for the aces since one of the aces must be from the same suit. This gives us 48 hands that become a straight flush. Similarly, there are 48 hands that become royal flushes. Thus, the 56,320 trip hands with A,A,A produce 48 royal flushes, 48 straight flushes, and 56,224 4-of-a-kinds.

There are 168,960 trip 6-card hands of the form A,x,x,x,y,z. The addition of a joker produces at least a full house. We again need to separate those that give straight flushes when a joker is added. If x,y,z is chosen from {2,3,4,5}, a straight flush may arise. There are 4 choices for the rank x, 3 choices for the other 2 ranks, 4 choices for the suit, and 3 choices for the other 2 cards of rank x. This yields 144 straight flushes. Similarly, we get 144 royal flushes. Hence, the 168,960 trip hands of the form under discussion produce 144 royal flushes, 144 straight flushes, and 168,672 full houses.

There are 506,880 trip 6-card hands not involving an ace. These hands may migrate to flushes, straights or stay the same. The number of ways of choosing 4 ranks without using A is . Each of these rank sets gives rise to 1,024 hands with trips. There are 9 sets with 4 consecutive ranks, there are 24 sets with a gap in rank that allows a straight upon adding a joker, and there are then 462 sets not allowing a straight upon adding a joker.

For the 9 sets with 4 consecutive ranks, there are 4 choices for a common suit, 4 choices for the rank of the trips, and 3 choices for the other 2 cards in the trips. This produces 48 hands with 4 suited cards, and they become straight flushes. The other 976 hands become straights. Thus, we get 48 royal flushes, 384 straight flushes, and 8,784 straights upon adding a joker to these 6-card hands.

For the 24 sets with a gap in the ranks allowing a straight when the joker is added, all 48 hands with 4 suited cards lead to straight flushes, but no longer is a royal flush possible. Thus, we obtain 1,152 straight flushes and 23,424 straights.

For the 462 rank sets with more than a single gap of 1, the addition of a joker cannot produce a straight. Thus, we multiply by 48 to obtain 22,176 hands that become flushes, and 450,912 hands that remain 3-of-a-kind.

Adding the various values for 6-card trip hands, we obtain contributions to Table 3 of

We move to the consideration of 2-pair 6-card hands. The number of ranks is at most 4 so that we may use much of what was done in the preceding case. There are 14,256 2-pair hands of the form A,A,x,x,y,y. Adding a joker turns these hands into full houses giving us 14,256 full houses.

There are 47,520 2-pair hands of the form x,x,y,y,z,z with no aces. Adding a joker leaves these hands as 2-pair giving us 47,520 2-pair hands.

There are 380,160 2-pair hands of the form A,A,x,x,y,z. The addition of a joker turns them into at least a full house. We must remove those that become straight flushes. There are 4 ways of choosing x,y,z from {2,3,4,5} and there are 3 ways of choosing which rank is to be paired. There are 4 choices for the common suit and 3 choices for the other A and the other x. This produces 432 straight flushes upon adding a joker. At the other end of the ranks, we obtain 432 royal flushes. Thus, these 6-card hands produce 432 royal flushes, 432 straight flushes, and 379,296 full houses.

This leaves us with the 2,090,880 6-card hands of the form x,x,y,y,z,w, where neither x nor y is A. A difference from the case of trips is that z or w could be A. So first let's look at the rank sets of cardinality 4 not containing an A, and use the information from the trips situation above.

Note that each rank set of cardinality 4 yields 3,456 2-pair hands as there are 6 choices for the ranks to be paired, 6 choices for each pair, and 4 choices for each card of the remaining 2 ranks. Of these 3,456 hands, the number with 4 suited cards is 216 because there are 6 choices for the ranks to be paired, 4 choices for the common suit, and 3 choices for each of the second cards of the rank that is paired.

First consider the 9 sets of 4 consecutive ranks. We obtain 1,728 straight flushes and 25,920 straights from the 8 that are not {10,J,Q,K}. The latter rank set is special in that adding a joker may produce a royal flush and a royal match. Of the 216 choices resulting in a royal flush, we have precisely 12 with 10,J,Q,K from 1 suit and K,Q from a second suit. Thus, these 9 rank sets produce 12 royal flushes with a royal match, 204 royal flushes, 1,728 straight flushes, and 29,160 straights.

The 24 sets with a single gap give 5,184 straight flushes and 77,760 straights.

The 462 other rank sets produce 99,792 flushes while leaving the remaining 1,496,880 hands as 2-pair.

We still must deal with the 220 rank sets containing A. Given that A cannot be one of the pairs, each set gives rise to 1,728 6-card hands of 2-pair. There are 108 6-card hands with 4 suited cards. Of the rank sets with an A, there are 4 allowing a small straight and 4 allowing a Broadway straight. Thus, we obtain 432 royal flushes, 432 straight flushes, 22,896 flushes, 12,960 straights, and 343,440 2-pair hands. Again, some of the choices producing royal flushes also have a royal match. They arise when the rank set is either {A,K,Q,J} or {A,K,Q,10}. Each allows 12 of these special hands. Thus, of the 432 royal flushes, 24 of them also have a royal match.

Adding the various values for 6-card 2-pair hands, we obtain contributions to Table 3 of

A 6-card hand with a single pair has exactly 5 ranks, but they cannot be consecutive or else the hand would be a straight. We know there are 1,277 sets of 5 ranks that are not consecutive. There are 7,680 possible 6-card hands for a given rank set of cardinality 5. Exactly 60 of them have 5 suited cards. This leaves 7,620 hands that are not already a flush. Of these, exactly 780 have 4 suited cards and become some kind of flush when a joker is added. This implies that for each subset of 4 ranks, there are 156 hands for which there are 4 suited cards of those ranks.

When considering royal flushes in the case of 6-card flushes and 5 distinct ranks, we saw that there are 39 rank sets allowing an A-high straight upon adding a joker. They do not all have the same behavior. The rank set {8,10,J,Q,K} produces 156 royal flushes, 156 straight flushes, 468 flushes, and 6,840 straights. When the 8 is replaced by any of the ranks from 2 through 7, the 156 straight flushes disappear and we have 624 flushes instead. The rank set {x,10,J,Q,A} produces 156 royal flushes, 156 straight flushes, 468 flushes, and 6,840 straights when x is either 8 or 9. For the other 6 values of x, we lose the 156 straight flushes. The rank sets {9,10,J,K,A}, {9,10,Q,K,A} and {9,J,Q,K,A} also gain an extra 156 straight flushes, while none of the others do. Thus, for these 39 rank sets, we have 6,084 royal flushes, 936 straight flushes, 23,400 flushes, and 266,760 straights.

For (x+4)-high straights, we again refer back to the subcase of 6-card flushes and 5 distinct ranks. The rank set {x,x+1,x+2,x+3} allows 6 choices for the other rank as long as x is not A. This gives us 156 (x+4)-high straight flushes, and either 624 flushes when the other rank is not x-2, or 312 straight flushes and 468 flushes when the other rank is x-2. For 5-high straights, we have 7 choices for the other rank, but we pick up no extra straight flushes. Thus, we have 55 rank sets here and 7 of them give an extra 156 straight flushes. Thus, we have 9,672 straight flushes, 33,228 flushes, and 376,200 straights.

For the rank sets {x,x+1,x+2,x+4}, we can use any of 7 other ranks. We pick up 156 straight flushes for each of the rank sets as well as an extra 156 straight flushes whenever the rank x-1 or x-2 is used. There are 63 rank sets and 15 of them allow an extra 156 straight flushes. This gives us 12,168 straight flushes, 36,972 flushes, and 430,920 straights.

For the rank sets {x,x+1,x+3,x+4}, we can use any of 7 other ranks. We pick up 156 straight flushes except when we use x-1 which gives an extra 156 straight flushes. There are then either 624 or 468 flushes, respectively. The rest of the hands produce straights. There are 63 rank sets like this and 8 with the rank x-1 included. Thus, we obtain 11,076 straight flushes, 38,064 flushes, and 430,920 straights.

For the rank sets {x,x+2,x+3,x+4}, we cannot use ranks x+5 or x+6 so that there are 6 other ranks for each, except when x is 9 and there are 7 other ranks. When x-1 is the other rank, we pick up an extra 156 straight flushes. Thus, there are 55 rank sets with 8 of them producing an extra 156 straight flushes. We obtain 9,828 straight flushes, 33,072 flushes, and 376,200 straights.

There are 10 sets of 5 ranks with 5 consecutive ranks, and there are 275 sets of 5 ranks, not all consecutive, that allow a straight upon adding a joker. There are then 1,002 rank sets that do not allow a straight upon adding a joker. Thus, such a rank set gives rise to hands that can improve to flushes, to 3 aces if there is a pair of aces, to 2-pairs if there is a pair and an ace, or remain a pair if the rank set does not contain an ace. Thus, we need to know how many of them contain an A as one of the ranks.

There are sets of 5 ranks containing A. Let's calculate how many of the rank sets considered above contain A. Of the 10 sets of 5 consecutive ranks, 2 contain A. Of the 39 rank sets producing A-high straights upon adding a joker, 32 contain A. Of the 55 rank sets containing {x,x+1,x+2,x+3}, 13 contain A. There are 14 for both {x,x+1,x+2,x+4} and {x,x+1,x+3,x+4}. The rank sets {x,x+2,x+3,x+4} have 12 containing an A. Thus, of the 285 rank sets already mentioned, 87 contain A. Therefore, we see that there are 408 rank sets with 5 elements containing A that do not allow a straight upon adding a joker, and there are 594 rank sets with 5 elements not containing an A that do not allow a straight upon adding a joker.

Consider the 408 rank sets containing an A that do not allow a straight upon adding a joker. There are 7,620 hands for a given rank set that do not have 5 suited cards. Each rank appears equally often as a pair, so that 1,524 of these hands have a pair of aces and the others do not. Of the 1,524 hands with a pair of aces, 156 have 4 suited cards. Hence, 156 of them become a flush upon adding a joker, while 1,368 become 3 aces. For the other hands, there are 624 flushes and 5,472 become 2-pairs. Thus, we obtain 318,240 flushes, 558,144 become 3 aces, and 2,232,576 become 2-pairs.

The 594 rank sets without an A either become flushes or remain as a single pair. This gives 463,320 flushes and 4,062,960 pairs.

Summing the above numbers for 6-card hands with a pair contributes

This now brings us to the 6-card hands that are simply high card as the best hand. The number of rank sets with 6 elements is . Nine of them have 6 consecutive ranks and 62 of them have 5 consecutive ranks but not 6. The remaining 1,645 rank sets contain at most 4 consecutive ranks. These are the sets with which we are concerned.

For a given rank set with 6 elements, there are possible hands. Exactly 4 of them have 6 suited cards and 72 of them have 5 suited cards. The remaining 4,020 hands give us high card hands and these are the hands in which we are interested.

Of the 4,020 possible hands giving high card hands, have 4 suited cards. They become some kind of flush upon adding a joker. Therefore, we obtain 888,300 flushes of some kind from the 1,645 rank sets. Since there are just 4 suited cards before adding the joker, the only special flushes that can be produced are royal flushes and straight flushes. We now determine the number of these special flushes. The essential observation to make is that each subset of 4 ranks among 6 given ranks is in precisely 36 flushes, with the given 4 ranks suited. This follows because there are 4 choices for the suit of the 4 ranks and 3 choices for the suit of the remaining ranks.

A set of 4 ranks producing a royal flush or a straight flush also allows straights. Thus, we shall be garnering information about straights as well.

In order to achieve Broadway, we must have one of the following rank sets: {J,Q,K,A}, {10,Q,K,A}, {10,J,K,A}, {10,J,Q,A} or {10,J,Q,K}. For {10,J,Q,K}, we are free to choose any 2 elements from {2,3,...,8} giving us 21 choices. For {10,J,Q,A}, we can choose any 2 elements from {2,3,...,9} except 8,9. Thus, there are 27 choices in this case. For any of the other 3 possible subsets, we can choose any 2 from 2 through 9 giving us 28 choices. Thus, we obtain 132 rank sets producing Broadway upon adding a joker.

For each of them, the choice of the 4 big ranks suited gives 36 royal flushes. Since this is the only way we can obtain royal flushes, there are 4,752 royal flushes coming from 6-card hands that are high card hands.

We choose to count the straight flushes via the 4 ranks that are suited. We first examine 4 ranks of the form x,x+1,x+2,x+3 that are suited, where x runs through A,2,...,9. We cannot have x+4 or x-1 in the rank set because that gives us 5 consecutive ranks. Note that rank x-1 is undefined when x is A. Thus, we are choosing 2 ranks from 7, which can be done in 21 ways, unless x is A, in which case we are choosing 2 ranks from 8, and this can be done in 28 ways. There are no further restrictions and this gives us 196 rank sets producing (x+4)-high straight flushes. This yields 7,056 straight flushes.

Consider the 4 ranks {x,x+1,x+2,x+4}. We are forbidden from using rank x+3, or both x-2,x-1 and that is it. Thus, for x = A and x = 2, we are choosing 2 elements from 8 which can be done in 28 ways. For x from 3 through 9, there are 27 choices because of excluding x-2,x-1. Altogether, we have 245 rank sets. This yields another 8,820 straight flushes.

We move to the set {x,x+1,x+3,x+4}. The only rank we are forbidden from using is x+2 since 2 distinct ranks chosen with these 4 ranks never produces 5 consecutive ranks. Hence, we are choosing 2 from 8 giving us 28 choices. The number of rank sets then is 252 leading to 9,072 straight flushes.

The last of the sets producing (x+4)-high straight flushes is {x,x+2,x+3,x+4}. In this case, we cannot use both ranks x+5 and x+6, as well as x+1. This restriction applies only to values of x from A,2,...,8. So we are choosing 2 elements from 8 when x is 8 or smaller with 1 choice not allowed. When x is 9, we are choosing 2 elements from 8. Hence, there are 244 of these sets altogether. This yields 8,784 straight flushes.

That takes care of all possible ways to get straight flushes. Summing the numbers gives 4,752 royal flushes and 33,732 straight flushes. This then leaves 849,816 flushes.

There are 778 rank sets of 6 ranks that do not contain 5 or more consecutive ranks but do allow a straight upon adding a joker. That leaves 867 rank sets of 6 ranks that do not allow a straight upon adding a joker. We first must determine how many of them contain A. If we go back and look through the 778 rank sets described above, we find there are 111 of the 132 sets allowing Broadway with an A in them. Of the 141 sets for {x,x+1,x+2,x+3}, 51 contain A. In the remaining cases, we obtain 62, 61 and 45, respectively. This gives 330 of these sets with an A in them.

Of the 9 sets with 6 consecutive ranks, precisely 2 have an A. Of the 62 sets with 5 consecutive elements, 20 contain A. There are of all possible rank sets with 6 elements containing an A. Thus, there are 440 sets containing an A that do not allow a straight upon adding a joker. For each of them, there are 540 hands producing a flush and 3,480 producing a pair of aces. This gives us 237,600 flushes and 1,531,200 pairs of aces.

We are left with 427 rank sets not containing an ace and not allowing a straight upon adding a joker. They produce 230,580 flushes and 1,485,960 high card hands.

We sum the various possible hands and obtain contributions of

We now sum all of the preceding values and obtain the numbers in the following table of possible Pai Gow bonus hands.

The above table of Pai Gow bonus hands lists the hands according to the rule that we choose the best possible 5-card poker hand, where we use the usual ranking for 5-card poker hands. However, the fact that the bonus payout for 3-of-a-kind exceeds that for a straight means that any 7-card hand with both a straight and 3-of-a-kind will be used as 3-of-a-kind for purposes of the bonus. Thus, we need to know how many hands have both a straight and a 3-of-a-kind. If there are X such hands, then we must subtract X from 11,236,028 and add X to 7,470,676 in order to count the hands regarding the bonus payouts.

We now determine the value of X. First we consider those without a joker. If the hand has both trips and a straight, then the rank set has the form x,x+1,x+2,x+3,x+4 and there are 10 such rank sets. There are 5 choices for the rank of the trips, 4 choices for the trips, and 4 choices for each of the other 4 cards. This gives us 5,120 hands for a given rank set. However, some of the hands are straight flushes. There are 5 choices for the rank of the trips, 4 choices for the trips and 3 choices for the common suit of the other 4 cards giving us 60 hands with a straight flush. Hence, we have 5,060 hands for a given rank set with a straight and trips. The 10 rank sets give us 50,600 7-card hands without a joker that have both a straight and trips in the hand.

Now let's look at the hands with a joker that contain a straight and trips. These hands arise in two possible ways. The 6 cards without the joker either have trips together with 3 other distinct ranks, where ace is not a rank that appears in the rank set, or there is a pair of aces together with 4 other distinct ranks. We consider each separately.

Earlier we saw that there are 32,208 3-of-a-kind hands not involving an ace that become straights. This is the contribution to X from 3-of-a-kind hands not involving an ace.

There are 87 sets of 5 ranks containing an ace that either already contain 5 consecutive ranks or allow a straight upon adding a joker. For each rank set, the total number of hands with a pair of aces is 1,536 as there are 6 choices for the pair and 4 choices for each of the other 4 cards. Of these 1,536 hands, 12 have 5 suited cards and 156 have 4 suited cards. This leaves 1,368 hands for each rank set not producing a flush or better. We obtain 119,016 hands that have trips and a straight.

Summing the above 3 numbers gives us X = 201,824. We then remove this number from the number of straights and add it to the number of trips. Therefore, the two lines of the above table for straights and 3-of-a-kind should be replaced by

We now present the number of hands according to the bonuses for which they are qualified.